Post on 19-Mar-2020
IONIC SOLIDS
https://www.youtube.com/w
atch?v=QXT4OVM4vXI
• Ions form when electrons are
exchanged between species (atoms or
groups of atoms)
• Ionic compounds form when these
charged ions attract each other
electrostatically
• The exist as large collections of the
bound ions and do not have a single
unit like molecules
What Is an Ionic Compound?
What Holds Ionics Together?
Representing in 2D and 3D...
Factors in Crystal Formation [ Fcoulomb = k(Q1Q2)r-2 ]
❏ Charges of Ions Involved…
❏ Relative Sizes of Ions Involved...
Electrical Conductivity - Ability to pass electrical current
Solid State
Explaining Ionic Macrostate Properties with
Microstate Structure
Molten OR Aqueous
High Melting Point...
∆Hdissociation = Elattice ∝ Q1Q2 / r
This means that the energy to break down
crystalline solids is related to two things…
• Charge of the ions involved (larger
charge greater attractions)
• The distance between the ions in the
lattice (closer equals greater attractions)
Explaining Ionic Macrostate Properties with
Microstate Structure
Nature of Gases
Gases are compressible, fill their containers, and essentially all behave similarly under normal
conditions
Kinetic Molecular Theory
1. Gases are composed of large numbers of particles that are in
constant, random motion
2. The volume of the particles is negligible compared to the total
volume they occupy.
3. Attractive and repulsive forces between gas particles are
negligible.
4. Energy can be transferred between molecules during collisions,
but the average kinetic energy of the molecules is constant with
time, i.e. all collisions are perfectly elastic.
5. The average kinetic energy of the particles is proportional to the
absolute temperature. At a given temperature the particles of
all gases have the same average kinetic energy!
The Gas Laws (Boyle’s) The PV product of a gas is equal to a constant value as
long as temperature and quantity (moles) are held
constant.
PV = constant
- SO -
P1V1 = P2 V2
Q: If 700mL of a gas at 2.5atm of pressure is expanded to
2.8L, what pressure does the gas exert?
A: 𝑃2 =𝑃1𝑉1
𝑉2=
2.5 𝑎𝑡𝑚 × 0.700 𝐿
2.8 𝐿
P2 = 0.63 atm
The Gas Laws (Charles’s Law) The V/T quotient of a gas is constant as long as volume and
quantity are held constant.
V = T constant
-SO -
V1 / T1 = V2 / T2
Q: At what temperature would the volume of a balloon
containing helium at 25C reach half its original value?
A: 𝑇2 =𝑇1
𝑉1× 𝑉2 =
298 𝐾
𝑉1×
1
2𝑉1
T2 = 149 K
The Gas Laws (Avogadro’s Law)
The V/n quotient of a gas is constant as long as temperature
and pressure are held constant
V = n constant
-SO -
V1 / n1 = V2 / n2
Q: What will be the new volume of a 5.6L nitrogen balloon at
STP if an additional 0.50mol of N2 is added?
A: 𝑉2 =𝑉1
𝑛1× 𝑛2 =
5.6 𝐿
𝑛1× 𝑛1 + 0.50 𝑚𝑜𝑙
𝑛1 = 5.6 𝐿 ×1 𝑚𝑜𝑙
22.4 𝐿= 0.25 𝑚𝑜𝑙
V2 = 16.8L
The Gas Laws (Gay-Lussac’s Law)
The P/T quotient of a gas is constant as long as the volume
and quantity of the gas are held constant
P = T constant
- SO -
P1 / T1 = P2 / T2
Q: At what temperature will the pressure of a gas initially at
20C and 700torr reach one third the initial pressure?
A: 𝑇2 =𝑇1
𝑃1× 𝑃2 =
293 𝐾
700 𝑡𝑜𝑟𝑟×
1
3× 700 𝑡𝑜𝑟𝑟
T2 = 97.7K
The Ideal Gas Law (BIG ONE)
Because the following things are known, we can describe
gases with one all-encompassing equation…
V 1 / P [Boyle]
V T [Charles]
V n [Avogadro]
- SO -
V nT / P , and R is defined as the proportionality constant
Which gives us V = R (nT / P)
- OR -
PV = nRT
The Ideal Gas Law
Q: Calcium carbonate, CaCo3 (s), decomposes when
heated to give CaO (s) and Co2 (g). A sample of CaCO3 is
decomposed, and the carbon dioxide produced is
collected in a 250mL flask. After the decomposition is
complete, the gas has a pressure of 1.3atm at a
temperature of 31C. How many moles of CO2 gas were
produced?
A: We are looking for moles and are not at standard
temperature and pressure…so use PV = nRT!
𝒏 =𝑷 × 𝑽
𝑹 × 𝑻=
𝟏. 𝟑 𝒂𝒕𝒎 × 𝟎. 𝟐𝟓𝟎 𝑳
𝟎. 𝟎𝟖𝟐𝟏 𝑳 𝒂𝒕𝒎 𝒎𝒐𝒍−𝟏𝑲−𝟏 × 𝟑𝟎𝟒 𝑲
n = 0.013 mol CO2 (g)
Special Uses of Ideal Law
Gas Density and Molar Mass
let M = molar mass
PV = nRT , (rearranging) … n/V = P/RT
nM = mass , density = d = mass / V
DENSITY = d = nM / V = PM / RT , so d = PM / RT
MOLAR MASS = M = dRT / P , M = dRT / P
Special Uses of Ideal Law
Q: Determine the density of xenon gas at 75C in a
container with a pressure of 2.00atm.
A: 𝒅 =𝑷×𝑴
𝑹×𝑻=
𝟐.𝟎𝟎 𝒂𝒕𝒎 × 𝟏𝟑𝟏.𝟑 𝒈 𝒎𝒐𝒍−𝟏
𝟎.𝟎𝟖𝟐𝟏 𝑳 𝒂𝒕𝒎𝒎𝒐𝒍−𝟏𝑲−𝟏 × 𝟑𝟒𝟖 𝑲
d = 9.18 g L1
Q: What is the molar mass of a gas with a density of
2.65g/L under the same conditions?
A: 𝑴 =𝒅×𝑹×𝑻
𝑷=
𝟐.𝟔𝟓 𝒈 𝑳−𝟏 × 𝟎.𝟎𝟖𝟐𝟏 𝑳 𝒂𝒕𝒎 𝒎𝒐𝒍−𝟏 𝑲−𝟏 × 𝟑𝟒𝟖 𝑲
𝟐.𝟎𝟎 𝒂𝒕𝒎
M = 37.9 g mol1
Daltons Law of Partial Pressure Because “ideal” gases behave independently of each other, they behave
very similarly. This means that a mix of gases will behave as several ideal
gases (under normal conditions) together, which means it will all behave
similarly…the only variable is the amount of each gas present in moles (n).
PT = P1 + P2 + … + Pn
Where Pn = nn (RT/V) or Pn = (nn / nT) PT
Q: What is the total pressure exerted by a mixture of 2.00g H2
and 8.00g N2 at 273K in a 10.0L vessel?
A: First, find moles of each…
2.00𝑔 𝐻2 ×1 𝑚𝑜𝑙 𝐻2
2.02 𝑔 𝐻2= 0.990 𝑚𝑜𝑙 𝐻2
8.00𝑔 𝑁2 ×1 𝑚𝑜𝑙 𝑁2
28.02 𝑔 𝑁2= 0.286 𝑚𝑜𝑙 𝑁2
then, find combined pressure…
𝑷𝒕𝒐𝒕𝒂𝒍 =𝒏𝒕𝒐𝒕𝒂𝒍𝑹𝑻
𝑽
𝑃𝑡𝑜𝑡𝑎𝑙 =0.99𝑚𝑜𝑙 + 0.286𝑚𝑜𝑙 8.31 𝐿 𝑘𝑃𝑎 𝑚𝑜𝑙−1𝐾−1 273𝐾
10.0 𝐿
Ptotal = 289 kPa
Partial Pressure
Collecting Gas Over Water
Collection of gas over water inherently adds water vapor to the mixture
and it must be accounted for via Dalton’s Law.
Q: A sample of KClO3 is partially decomposed producing O2
gas that is collected over water. The volume of gas
collected is 0.250L at 26C and 765torr total pressure.
How many moles of O2 were collected? What mass of
KClO3 decomposed?
A: 1) Determine pressure of O2…
𝑃𝑂2= 𝑃𝑡𝑜𝑡𝑎𝑙 − 𝑃𝐻2𝑂 = 765 𝑡𝑜𝑟𝑟 − 25.2 𝑡𝑜𝑟𝑟 = 740 𝑡𝑜𝑟𝑟 = 0.974 𝑎𝑡𝑚
2) Determine moles of O2…
𝑛𝑂2=
𝑃𝑂2𝑉
𝑅𝑇=
0.974 𝑎𝑡𝑚 0.250 𝐿
0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 299 𝐾= 9.92 × 10−3 𝑚𝑜𝑙
3) Use stoichiometry to find masss of KClO3 …
9.92 × 10−3 𝑚𝑜𝑙 𝑂2 ×2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3
3 𝑚𝑜𝑙 𝑂2×
122.5 𝑔 𝐾𝐶𝑙𝑂3
1 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3= 0.810 𝑔 𝐾𝐶𝑙𝑂3
BCE: 2KClO3 2KCl + 3O2
KEavg and urms
KEavg = 3/2 RT
Relationship of temperature and KE
urms = [3RT/M]1/2
M must be in kg in order to work
out…represents average speed, and u is not
the average velocity…but is close
Effusion and Diffusion
Both concepts rely heavily on the random motion of
gases…basically they relate the fact that while kinetic energy is
the same for molecules at the same temperature, the root-
mean-square velocity for lighter molecules is higher because
they have a lower mass.
EFFUSION: 𝑟𝑎𝑡𝑒2
𝑟𝑎𝑡𝑒1=
𝑀1
𝑀2
DIFFUSION: 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒2
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒1=
𝑀1
𝑀2
Escape of a
gas through
a small hole
Spreading out
of a gas in space
van der Waals Equation
𝑃 = 𝑛𝑅𝑇
𝑉 − 𝑛𝑏−
𝑛2𝑎
𝑉2
Adjusts for the fact that
atoms DO have volume!
Adjusts for intermolecular
attractions!
Greatest deviations at HIGH PRESSURE and LOW
TEMPERATURE!